hw7_sol

hw7_sol - ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F....

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Unformatted text preview: ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-114* Determine the magnitude and direction of the resultant of the two forces shown in Fig. P4~114 and the perpendicular distance dR from point A to the line of action of the resultant. F¢.P4414 SOLUTION R = = 300 cos 45° + 480 cos 30° = 981.38 N X = 800 sin 45° - 480 sin 30° = 325.69 N I 2 2 / , 2 2 x Ry = (981.38) + (325.69) = 1034.01 N a 1034 N R -1 981.38 cos" 51 = cos ——————— = 18.3590 2 18.360 1034.01 = 1034 N a 18.360 300 sin 45° (0.320) — 800 cos 45° (0.120) 400 cos 30° (0.240) - 400 sin 30°(0.640) —140.23 N-m = 140.23 N-n D RdR = 1034.0dR = 140.23 N'm 0.13562 m § 135.6 mm 23‘343 ENGINEERING MECHANICS - STATICS, 2nd. F. RILEY AND L. D. STURGES 4-115 Replace the three forces Fh=lﬂow Shown in Fig. P4—115 by an equivalent force-couple E :3 P! system at point B. FD=3me Fig. 134-115 SOLUTION -100 lb 200 - 300 = -100 lb {-100}2 + {—10012 = 141.42 1b 3 141.4 lb R x -1 -100 _ _ O _ - E- w 003 141_42 — 133.00 — 130.0 —1 COS 0 = 141.4 lb y 45.0° C = 2MB = 100(10) - 300(14} + 200(20} = 800 in.-lb a 800 1n.-1b j 22‘9'7 ENGINEERING MECHANICS - STATICS, 2nd. Replaca the three forces shown in Fig. P4—116 by an equivalent force—couple system at point 0. SOLUTION = BF X — -400 + %(750} = 200 N 2F _ §(?50) + 100 = Y D 550 N Ed. W. F. RILEY AND L. D. STURGES (200}2 + (550}2 = 585.2 N e 535 N -‘1 -1 200 COS .1 _ _ R ‘ COS 585.2 ' = 585 N a 7o.0° EM = 400(0.160) + 100(U.360} EZ‘IiB ?0.02° e 70.0° 100.0 N'm Fig. P4416 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 4-11?‘ Three forces are applied to 3. been as shown in In“: 15m Wlb Fig. Pd-IIT. Deter-ine the resultant R DI the three forces and the location 0f its line of action with respect tn Euppﬂrt A” Fig. P441? SOLUTION R 95 3 1b = 95 1h T + c Ema 30(14: - ?5{3ﬂ1 + 9n:4n1 = 24TD in.-lh = 24TD in.-1h 3 EH“. = Rx“ = Eﬁxﬂ = 2415' innlb 1“ = 25.0 in. -1 ENGINEERING MECHANICS - STATICS W. F. RILEY AND L. D. STURGES 4-124* Four forces are applied to a post as shown in Fig. P4-124. Determine the resultant E of the four forces and the location of its line of action with respect to support E. Fig. P4-1 24 SOLUTION cos 25° - 90 cos 40° — 75 + 60 = -11.44 N sin 25° + 90 sin 40° = 91.66 N /(—11.44)2 + (91.66)2 = 92.37 N = 92.4 N ‘1 _11144 92.37 0 - cos = 97.11° a 97.1 92‘4 N s 82.90 -80 cos 25° (1000) + 90 cos 400 (680) 75(440) — 60(240) = -7023 N'mm 7023 N‘mm 3 _ IIE - .19§§ — 76 0 m R ‘ 92.37 ’ ' ” ENGINEERING MECHANICS - STATICS W. F. RILEY AND L. D. STURGES Four forces and a couple are applied to a frame as shown in Fig. P4—126. Determine the magnitude and direction of the resultant and the perpendicular distance dR from point A to the line of action of the resultant. Fig. P4-126 SOLUTION 16_ o 30 — 28.07 cos 28.070 — 50 - 75 = —54.41 lb sin 28.o7° — 90 = —52.36 lb = y’(-54.41)2 + (-52.36)2 = 75.51 lb 2 75.5 lb ‘1 '54041 S 0 O 75.51 — -136.10 — -136.1 CO 75.5 lb y 43.9° 75(12) + 50(30) — 100 - 80 cos 28.070 (30 — 10 cos 28.090) + 80 sin 28.07° (16 + 10 sin 28.070) - 90(16 + 18 tan 28.090 + 14) —1979.3 in.°lb = 1979.3 in.-lb D — 26.21 in. 3 26.2 in. 257 ...
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hw7_sol - ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F....

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